Question

The sum of intercepts on co-ordinate axes made by tangent to the curve $\sqrt{x} + \sqrt{y} = \sqrt{a}$ is

• A. a
• B. 2a
• C. $2\sqrt{a}$
• D. None of these

Answer 1

Answer: (A)

a

Answer 2

$\sqrt{x} + \sqrt{y} = \sqrt{a}$ ⇒ $\frac{1}{2\sqrt{x}} + \frac{1}{2\sqrt{y}}\frac{dy}{dx} = 0$ ⇒ $\therefore$ $\frac{dy}{dx} = - \frac{\sqrt{y}}{\sqrt{x}}$

Hence tangent at (x, y) is $Y - y = - \frac{\sqrt{y}}{\sqrt{x}}(X - x)$ or

$X\sqrt{y} + Y\sqrt{x} = \sqrt{xy}(\sqrt{x} + \sqrt{y}) = \sqrt{axy}$ or $\frac{X}{\sqrt{a}\sqrt{x}} + \frac{Y}{\sqrt{a}\sqrt{y}} = 1$

Clearly its intercepts on the axes are $\sqrt{a}\sqrt{x}$ and $\sqrt{a}\sqrt{y}$.

Sum of the intercepts = $\sqrt{a}(\sqrt{x} + \sqrt{y}) = \sqrt{a}.\sqrt{a} = a$.